sábado, 29 de agosto de 2020

ecuacions diferencials y producte integral

( y(x) )^{2} [o(x)o] d_{x}[y(x)] = f(x)

y(x) = ( 3·int[f(x)] d[x] )^{(1/3)}

( 3·int[f(x)] d[x] )^{(2/3)} [o(x)o]...

f(x) / [o(x)o] / ( 3·int[f(x)] d[x] )^{(2/3)} = f(x)


( y(x) )^{n} [o(x)o] d_{x}[y(x)] = f(x)

y(x) = ( (n+1)·int[f(x)] d[x] )^{(1/(n+1))}

( (n+1)·int[f(x)] d[x] )^{(n/(n+1))} [o(x)o]...

f(x) / [o(x)o] / ( (n+1)·int[f(x)] d[x] )^{(n/(n+1))} = f(x)


d_{x}[f(x) / [o(x)o] / ( 3·int[f(x)] d[x] )^{(2/3)}] = ...

... ( ( 3·int[f(x)] d[x] )^{(1/3)}/f(x) )·d_{x}[f(x)]

d_{x}[f(x) / [o(x)o] / ( (n+1)·int[f(x)] d[x] )^{(n/(n+1))}] = ...

... ( ( (n+1)·int[f(x)] d[x] )^{(1/(n+1))}/f(x) )·d_{x}[f(x)]


d_{x}[ d_{x}[g(x)] / [o(x)o] / ( f(x) )^{n} ] = ...

... ( ( 1/((-n)+1) )·( f(x) )^{(-n)+1}/d_{x}[f(x)] )·d_{xx}^{2}[g(x)]


d_{x}[ ( ( 1/((-n)+1) )·( f(x) )^{(-n)+1} / [o(x)o] / d_{x}[f(x)] ) [o(x)o] d_{xx}^{2}[g(x)] ] = ...

... ( ( 1/((-n)+1) )·( f(x) )^{(-n)+1}/d_{xx}^{2}[f(x)] )·d_{xxx}^{3}[g(x)]

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